## MATH1901 Quizzes

Quiz 10: Partial derivatives and tangent planes
Question 1 Questions
Which option correctly gives the two first order partial derivatives of the following function?
$f\left(x,y\right)={e}^{x}+\frac{x}{y}+{\left(2x+y\right)}^{4}$
Exactly one option must be correct)
 a) ${f}_{x}={e}^{x}+\frac{1}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=xlny+4{\left(2x+y\right)}^{3}$ b) ${f}_{x}={e}^{x}+\frac{{x}^{2}}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=-\frac{x}{{y}^{2}}+8{\left(2x+y\right)}^{3}$ c) ${f}_{x}={e}^{x}+\frac{{x}^{2}}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=xlny+8{\left(2x+y\right)}^{3}$ d) ${f}_{x}={e}^{x}+\frac{1}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=-\frac{x}{{y}^{2}}+4{\left(2x+y\right)}^{3}$ e) ${f}_{x}={e}^{x}+\frac{{x}^{2}}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}={e}^{x}-\frac{x}{{y}^{2}}+8{\left(2x+y\right)}^{3}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
Find the two first order partial derivatives, with respect to $x$ and $y$, of
$z=cos\left({x}^{2}y\right)+siny.$
Exactly one option must be correct)
 a) $\frac{\partial z}{\partial x}=-2xsin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}=-{x}^{2}ysin\left({x}^{2}y\right)-cosy$ b) $\frac{\partial z}{\partial x}=-2xysin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}={x}^{2}sin\left({x}^{2}y\right)-cosy$ c) $\frac{\partial z}{\partial x}=2xsin\left({x}^{2}y\right)-cosy$ and $\frac{\partial z}{\partial y}={x}^{2}sin\left({x}^{2}y\right)$ d) $\frac{\partial z}{\partial x}=-2xysin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}=-{x}^{2}sin\left({x}^{2}y\right)+cosy$ e) $\frac{\partial z}{\partial x}=-{x}^{2}sin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}=-2xsin\left({x}^{2}y\right)+cosy$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
Find the first order partial derivative with respect to $y$ of
$f\left(x,y\right)=x{e}^{x{y}^{2}}+log\left(xy\right).$
Exactly one option must be correct)
 a) ${e}^{x{y}^{2}}+2y{x}^{2}{e}^{x{y}^{2}}+\frac{1}{y}+\frac{1}{x}$ b) $2y{x}^{2}{e}^{x{y}^{2}}+\frac{1}{y}$ c) $x{e}^{x{y}^{2}}+\frac{1}{xy}$ d) ${e}^{x{y}^{2}}+{y}^{2}x{e}^{x{y}^{2}}+\frac{1}{x}$ e) $2y{x}^{2}{e}^{x{y}^{2}}+\frac{1}{xy}$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Find the first order partial derivative with respect to $x$ of
$f\left(x,y\right)=\frac{1}{{x}^{2}+{y}^{2}}.$
Exactly one option must be correct)
 a) $2xlog\left({x}^{2}+{y}^{2}\right)$ b) $-\frac{2y}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$ c) $-\frac{2x}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$ d) $2ylog\left({x}^{2}+{y}^{2}\right)$ e) $-\frac{1}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Find the tangent plane to the surface
$z={x}^{2}-{y}^{2}$
at the point (5,-4,9). Exactly one option must be correct)
 a) $z-9=10\left(x-5\right)+8\left(y+4\right)$ b) $z-9=10\left(x-5\right)-8\left(y-4\right)$ c) $z-9=8\left(x+4\right)+10\left(y-5\right)$ d) $z-9=8\left(x-5\right)-10\left(y-4\right)$ e) $z-9=5\left(x-8\right)+4\left(y-8\right)$

Choice (a) is correct!
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
The differential of the function
$z=cos\left(\frac{{x}^{2}}{x+y}\right)$
is given by Exactly one option must be correct)
 a) $\frac{dz}{dx}=-\frac{{x}^{2}+2xy}{{\left(x+y\right)}^{2}}sin\left(\frac{{x}^{2}}{x+y}\right)$. b) $dz=-\frac{{x}^{2}+2xy}{{\left(x+y\right)}^{2}}sin\left(\frac{{x}^{2}}{x+y}\right)dx+\frac{{x}^{2}}{{\left(x+y\right)}^{2}}cos\left(\frac{{x}^{2}}{x+y}\right)dy$. c) $dz=-\frac{{x}^{2}+2xy}{{\left(x+y\right)}^{2}}sin\left(\frac{{x}^{2}}{x+y}\right)dx+\frac{{x}^{2}}{{\left(x+y\right)}^{2}}sin\left(\frac{{x}^{2}}{x+y}\right)dy$. d) $dz=-\frac{{x}^{2}+2xy}{{\left(x+y\right)}^{2}}sin\left(\frac{{x}^{2}}{x+y}\right)dy-\frac{{x}^{2}}{{\left(x+y\right)}^{2}}sin\left(\frac{{x}^{2}}{x+y}\right)dx$. e) $dz=z-{z}_{0}=cos\left(\frac{{x}^{2}}{x+y}\right)-cos\left(\frac{{x}_{0}^{2}}{{x}_{0}+{y}_{0}}\right)$. f) None of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Choice (f) is incorrect
If $f\left(-1,3\right)=4,\phantom{\rule{1em}{0ex}}{f}_{x}\left(-1,3\right)=5$ and ${f}_{y}\left(-1,3\right)=-2$, the linear approximation to $f\left(-1.3,3.2\right)$ is given by: Exactly one option must be correct)
 a) 2.1 b) 4 c) 5.9 d) 4.1 e) there is not enough information given to estimate $f\left(-1.3,3.2\right)$.

Choice (a) is correct!
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Using differentials, and without using a calculator, the approximate value of $f\left(x,y\right)={x}^{2}{e}^{3x-y}$ at (2.25,5.5) is Exactly one option must be correct)
 a) 2 b) 4 c) 10 d) 18 e) 27

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Use $f\left(x+dx,y+dy\right)=f\left(x,y\right)+{f}_{x}\left(x,y\right)dx+{f}_{y}\left(x,y\right)dy$ with $x=2,\phantom{\rule{1em}{0ex}}y=6,\phantom{\rule{1em}{0ex}}dx=0.25,\phantom{\rule{1em}{0ex}}dy=-0.5$. In this way ${e}^{3x-y}={e}^{0}$ and hence $f\left(2,6\right),\phantom{\rule{1em}{0ex}}{f}_{x}\left(2,6\right)$ and ${f}_{y}\left(2,6\right)$ can all be evaluated without a calculator. Actually response 4 is closer to the exact value but is not what is obtained when this method of approximation is used.
Choice (d) is incorrect
Choice (e) is incorrect
Using differentials, the approximate volume of tin in a closed cylindrical tin can with radius 4cm, height 12cm and where the thickness of metal is 0.04cm, is Exactly one option must be correct)
 a) $3.84\pi \phantom{\rule{1em}{0ex}}{\text{cm}}^{3}$. b) $5.12\pi \phantom{\rule{1em}{0ex}}{\text{cm}}^{3}$. c) $24.8\pi \phantom{\rule{1em}{0ex}}{\text{cm}}^{3}$. d) $16.226\phantom{\rule{1em}{0ex}}{\text{cm}}^{3}$. e) $12.8\pi \phantom{\rule{1em}{0ex}}{\text{cm}}^{3}$.

Choice (a) is incorrect
Choice (b) is correct!
The volume of the tin is the difference between the outside volume and the inside volume of the tin which is approximately equal to the total differential of the inside volume. The volume of a cyclinder of height $h$ and radius $r$ is $V=\pi {r}^{2}h$. Hence, the differential is
$dV=\frac{\partial V}{\partial r}dr+\frac{\partial V}{\partial h}dh=2\pi rh\phantom{\rule{0.3em}{0ex}}dr+\pi {r}^{2}\phantom{\rule{0.3em}{0ex}}dh.$
Now, $dr=0.04$ and $dh=0.08$ (Note that the cylinder is closed and so the $dh=0.08$ comes from the contribution of the top and bottom of the cylinder which together are twice the thickness of the metal.) Therefore,
$dV=2\pi \cdot 4\cdot 12\cdot 0.04+\pi \cdot {4}^{2}\cdot 0.08=5.12\pi \phantom{\rule{1em}{0ex}}{\text{cm}}^{3}.$
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
The volume of a circular cylinder is given by $V=\pi {r}^{2}h$ where $r$ is the radius of the cylinder and $h$ is its height. A circular cylinder with $r=3$cm and $h=2$cm has a volume of $18\pi$ cubic centimetres. The radius of the cylinder is now reduced by 0.5cm and the height by 0.2cm. An engineer makes an estimate of the reduction in volume using differentials. What is this estimate? Exactly one option must be correct)
 a) $6.75\pi$ cubic centimetres b) $6.9\pi$ cubic centimetres c) $7.8\pi$ cubic centimetres d) $12.6\pi$ cubic centimetres e) 21.21 cubic centimetres

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect